[Math] X follows an exponential distribution, calculate Expected value of sqrt(X).

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Problem: Let X follow an exponential distribution with expected value of 1. Define Y=sqrt(X). Calculate E(Y).

This is my first course in probability theory (5 weeks ≈ about 5*40 hours of workload) so the tools we have learned are not that many.

Tip: if X follows N(0,1) then E(X^2)=1.

Attempt: enter image description here

Correct answer it sqrt(Pi/2)

Best Answer

An exponential distribution of a random variable $X$ of expected value $1$ (i.e., mean) has PDF $f_X(x) = e^{-x}$ for $x \gt 0$, and zero otherwise. You want $E(\sqrt{X})$.

In general, to compute $E[(g(X)]$ for a distribution $f_X(x)$ is

$$\int_{-\infty}^{\infty} dx \, g(x) f(x) $$

In your case, the integral is

$$\int_0^{\infty} dx \, \sqrt{x} \, e^{-x}$$

To evaluate the integral, sub $x=y^2$ and get

$$2 \int_0^{\infty} dy \, y^2 \, e^{-y^2} = \int_{-\infty}^{\infty} dy \, y^2 e^{-y^2}$$

The result is, in fact $\sqrt{\pi}/2$.

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